<span>We know , E = kQ/r^2 where q = charge and r is separation between point and point charge.
Now, At P, E= kQ/r^2
Since, Q can't be changed, we can do that by varying r
2E = 2kq/r^2
2E = kq/ (r/ sqrt2)^2
Hence, if we bring Q closer such that distance between P and Q becomes r/ sqrt 2, E will get doubled.</span>
Answer:
a_total = 2 √ (α² + w⁴)
, a_total = 2,236 m
Explanation:
The total acceleration of a body, if we use the Pythagorean theorem is
a_total² = a_T²2 + 
²
where
the centripetal acceleration is
a_{c} = v² / r = w r²
tangential acceleration
a_T = dv / dt
angular and linear acceleration are related
a_T = α r
we substitute in the first equation
a_total = √ [(α r)² + (w r² )²]
a_total = 2 √ (α² + w⁴)
Let's find the angular velocity for t = 2 s if we start from rest wo = 0
w = w₀ + α t
w = 0 + 1.0 2
w = 2.0rad / s
we substitute
a_total = r √(1² + 2²) = r √5
a_total = r 2,236
In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m
a_total = 2,236 m
The frictional force is directly proportional to the force that is perpendicular on the surface.
When the body is placed on a horizontal level with zero inclination, the only force acting on the body is the gravitational force which always pulls the body down. The gravitational force, in this case, is the perpendicular force to the surface. Accordingly, this entire force is used to generate friction
Now as the inclination of the surface increases, the gravitational force is no longer the perpendicular force of the body, its value decreases, which means only a part is used to generate frictional force. Consequently, frictional force decreases.
When the inclination reaches 90 degrees, the gravitational force does not act along the normal and accordingly, no friction force is generated.
Therefore, elements whose atoms can have the same number of valence electrons are grouped together in the periodic table of the elements. As a general rule, a main group element (except hydrogen or helium) tends to react to form a closed shell, corresponding to the electron configuration s2p6.
Hoped I Helped
Answer:
The speed of the clay before the impact was 106.35 m/s.
Explanation:
the only force doing work on the system is the frictional force, f, the work done by f is given by:
Wf = ΔK = Kf - Ki
The clay and the block will come to rest after sliding Δx = 7.50 m, if their intial speed is v and the combined mass is m and μ is the coefficient of friction and g is gravity,then:
f×Δx = Ki
m×g×Δx×μ = 1/2×m×v^2
v^2 = 2×g×Δx×μ
= 2×(9.8)×(7.50)×(0.650)
= 95.55
v = 9.78 m/s
This is the veloty of clay and block after the clay hit the block.
if the velocity the clay and block attains after the impact is v and the initial speed of the clay is v1 and the mass is m and the speed of the block initially is V = 0 m/s and the mass is M, then according to the conservation of linear momentum:
m×v1 +M×V = v(m + M)
m×v1 = v(m + M)
v1 = v(m + M)/m
v1 = (9.78)(8.3×10^-3 + 82×10^-3)/(8.3×10^-3)
v1 = 106.35 m/s
Therefore, the speed of the clay before the impact was 106.35 m/s.
